If vectors s and b are orthogonal then, ||2s − 3b||^2 = 4||s||^2 + 9||b||^2. True or False and explain. I have no idea how to start this problem, I know, b dot s = 0 therefore, sin(theta)||b||s|| = 0 Not sure how i can apply this to the problem though.
I would try use the fact that ||v||^2 = v dot v to expand both sides. See what happens
looks true to me, just use the fact that \(\large a^2 = a \bullet a\)
oh damn, I cant believe I missed that property ugh
apply the definition of distance and inner product ||a||^2 = (a,a)
so that ||2s-3b||^2 =(2s-3b, 2s-3b)
it's easy now, right? can you continue?
I dont think I follow actually ugh
ok, ((2s-3b),(2s-3b)) left distribute =((2s-3b),2s)-((2s-3b),3b)
left distribute again ((2s,2s)-(3b,2s)-(2s,3b)+(3b,3b)
the 2middle term =0 because s \(\perp\)b so that you just have (2s,2s)+(3b,3b)
by definition (2s,2s) =||2s||^2 and (3b,3b)=||3b||^2 done
one more thing, take constants out to get your form 4||s||^2+9||b||^2
wait I think I see, ||2s - 3b||^2 = (2s - 3b) ∙ (2s - 3b) = (2s - 3b) ∙ (2s - 3b) = 2s∙2s - 2s∙3b - 3b∙2s + 3b∙3b = 2s∙2s - 2(2s∙3b) + 3b ∙ 3b oh ok I get it thanks
if they weren't orthogonal -2(2s∙3b) =/= 0 therefore s and b must be orthogonal to be equal
Ugh I need to study the rules harder thanks loser for the assistance
I don't know why you use \(\bullet\) , it is not the correct notation. However, If you are familiar with this, it 's ok
I dont see how it is incorrect it is the dot product and that is a dot?
we use comma.
throughout my textbook it is a dot and in my notes so I dont know maybe your way is correct, does it really matter that much?
I don't know, when I took elementary linear algebra, my prof used dot, too. But when I took advance linear, the prof use comma and all reference books use comma. So that I use it only.
gtg, see you later.
That is pretty funny, I dont know what to tell you. Maybe it is just to separate the men from boys.
I am sorry for leaving before you finished. My food burned...... hehehe.. that's the reason.
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